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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Aqualim on December 03, 2009, 01:57:13 pm

Title: Some silly Question
Post by: Aqualim on December 03, 2009, 01:57:13 pm: Just looking back over the Year 11 book to see if there was anything i didn't know how to do and stumbled along this:

Find the acute angle between the lines with equations $y=2x+6$ and $y=5-4x$
Title: Re: Some silly Question
Post by: Ilovemathsmeth on December 03, 2009, 02:28:35 pm: Find the gradient of each line; that is, 2 and -4 respectively.

Keep your calc in degree mode.

Find tan inverse (2) = 63.43494882 deg.
Find tan inverse (-4) = -75.96375653; add 180 to give 104.0362435 deg.

The rule theta 2 -theta 1 gives interior angle; i.e. larger angle - smaller angle.

Answer should be 104.0362435 - 63.43494882 = 40.60129465

Watch out if it asks 'obtuse angle' - you'd need to subtract above angle from 180 if so.

Hope that helps.
Title: Re: Some silly Question
Post by: Ilovemathsmeth on December 03, 2009, 02:29:48 pm: The reason why you add 180 above is because you want to find the angle each line makes with the positive direction of the x-axis. In the above, the neg result is not in the pos direction.
Title: Re: Some silly Question
Post by: Aqualim on December 03, 2009, 02:38:21 pm: Cheers thanks for that, I was thinking something along the $m=\tan \theta$ line.. So when you sketching it, how do you know which part you are look for, as in which part of the graph are you looking for the distance between
Title: Re: Some silly Question
Post by: qshyrn on December 03, 2009, 04:11:19 pm: heres how i did it (look at attachment) hope u understand
EDIT: umm u dont really need to find out which one is the obtuse/acute at first, just find one of the angles between the lines then u can find out the other one and therefore know which one is obtuse and acute
Title: Re: Some silly Question
Post by: /0 on December 03, 2009, 08:55:05 pm: Another way for people doing spesh is to turn them into vectors:

A vector representing $y=2x+6$ is $\mathbf{a} = \mathbf{i}+2\mathbf{j}$

A vector representing $y = 5-4x$ is $\mathbf{b} = \mathbf{i}-4\mathbf{j}$

Thus, by the dot product, we have

$\cos{\theta} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}= \frac{(1)(1)+(2)(-4)}{\sqrt{1^2+2^2}\sqrt{1^2+(-4)^2}}=-\frac{7\sqrt{85}}{85}$

So the angle between the vectors is $139.4^{\circ}$ , and the acute angle is $180 - 139.4 = 40.6^{\circ}$
Title: Re: Some silly Question
Post by: crappy on December 03, 2009, 08:56:03 pm: Quote from: /0 on December 03, 2009, 08:55:05 pm
Another way for people doing spesh is to turn them into vectors:

A vector representing $y=2x+6$ is $\mathbf{a} = \mathbf{i}+2\mathbf{j}$

A vector representing $y = 5-4x$ is $\mathbf{b} = \mathbf{i}-4\mathbf{j}$

Thus, by the dot product, we have

$\cos{\theta} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}= \frac{(1)(1)+(2)(-4)}{\sqrt{1^2+2^2}\sqrt{1^2+(-4)^2}}=-\frac{7\sqrt{85}}{85}$

So the angle between the vectors is $139.4^{\circ}$ , and the acute angle is $180 - 139.4 = 40.6^{\circ}$

gotta love spesh lol
Title: Re: Some silly Question
Post by: /0 on December 03, 2009, 08:57:29 pm: haha yeah I never got the hang of angles between lines. I trust vectors though :)
Title: Re: Some silly Question
Post by: Aqualim on December 04, 2009, 11:03:10 am: So basically in a nutshell, if they ask you to find the Acute angle, you add 180 (to ensure that the figure is a positive number), but for obtuse you would subtract 180 make the figure negative??
Title: Re: Some silly Question
Post by: Ilovemathsmeth on December 04, 2009, 11:26:14 am: Amazing, /0, you turned those lines into vectors so quickly! Are vectors hard in Spesh?
Title: Re: Some silly Question
Post by: Hooligan on December 04, 2009, 09:15:14 pm: Quote from: Ilovemathsmeth on December 04, 2009, 11:26:14 am
Amazing, /0, you turned those lines into vectors so quickly! Are vectors hard in Spesh?

Nahhh... vectors aren't too hard. xD You aced Methods, so I'm pretty sure you wouldn't find Spesh too hard. :)
Title: Re: Some silly Question
Post by: Hooligan on December 04, 2009, 09:16:27 pm: Quote from: /0 on December 03, 2009, 08:55:05 pm
Another way for people doing spesh is to turn them into vectors:

A vector representing $y=2x+6$ is $\mathbf{a} = \mathbf{i}+2\mathbf{j}$

A vector representing $y = 5-4x$ is $\mathbf{b} = \mathbf{i}-4\mathbf{j}$

Thus, by the dot product, we have

$\cos{\theta} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}= \frac{(1)(1)+(2)(-4)}{\sqrt{1^2+2^2}\sqrt{1^2+(-4)^2}}=-\frac{7\sqrt{85}}{85}$

So the angle between the vectors is $139.4^{\circ}$ , and the acute angle is $180 - 139.4 = 40.6^{\circ}$

Hey /0, can you show me how you converted those linear equations into vector form? I'm interested... I never recall converting linear to vector in Spesh...
Title: Re: Some silly Question
Post by: /0 on December 04, 2009, 09:35:26 pm: I just picked vectors that had the same 'gradient' as the lines.
Since the location of the vector isn't important, I ignored the $+6$ and $5$ translations.

Instead I had $y=2x$ and $y=-4x$

Since the coefficient of $\mathbf{j}$ gives the 'rise' and the coefficient of $\mathbf{i}$ gives the 'run',

You can say $\mathbf{r} = x\mathbf{i}+y\mathbf{j}$

And then sub in
Title: Re: Some silly Question
Post by: Ilovemathsmeth on December 05, 2009, 12:45:56 am: Vectors sound exciting but a little scary, so many weird lines in the Spesh book :P

Thanks Hooligan :) I'm so lazy, I'll have to start it soon but I've got ages, especially if I take Gap year.
Title: Re: Some silly Question
Post by: Hooligan on December 11, 2009, 04:00:03 pm: Quote from: /0 on December 04, 2009, 09:35:26 pm
I just picked vectors that had the same 'gradient' as the lines.
Since the location of the vector isn't important, I ignored the $+6$ and $5$ translations.

Instead I had $y=2x$ and $y=-4x$

Since the coefficient of $\mathbf{j}$ gives the 'rise' and the coefficient of $\mathbf{i}$ gives the 'run',

You can say $\mathbf{r} = x\mathbf{i}+y\mathbf{j}$

And then sub in

How interesting! ^_^ Thanks /0!! I have never used Vectors in this sort of setting. :)
Title: Re: Some silly Question
Post by: Hooligan on December 11, 2009, 04:01:45 pm: Quote from: Ilovemathsmeth on December 05, 2009, 12:45:56 am
Vectors sound exciting but a little scary, so many weird lines in the Spesh book :P

Thanks Hooligan :) I'm so lazy, I'll have to start it soon but I've got ages, especially if I take Gap year.

Hehehe.... Spesh always seems to be daunting, but actually is just a different part of Maths, and just so happens to be 'hard' as most perceive it. :P
Title: Re: Some silly Question
Post by: Gloamglozer on December 11, 2009, 04:59:30 pm: Quote from: Hooligan on December 11, 2009, 04:01:45 pm
Quote from: Ilovemathsmeth on December 05, 2009, 12:45:56 am
Vectors sound exciting but a little scary, so many weird lines in the Spesh book :P

Thanks Hooligan :) I'm so lazy, I'll have to start it soon but I've got ages, especially if I take Gap year.

Hehehe.... Spesh always seems to be daunting, but actually is just a different part of Maths, and just so happens to be 'hard' as most perceive it. :P

After having a look at the Heinemann textbook, I think it's very physic-sy. :)
Title: Re: Some silly Question
Post by: Hooligan on December 11, 2009, 06:30:30 pm: Quote from: Gloamglozer on December 11, 2009, 04:59:30 pm
Quote from: Hooligan on December 11, 2009, 04:01:45 pm
Quote from: Ilovemathsmeth on December 05, 2009, 12:45:56 am
Vectors sound exciting but a little scary, so many weird lines in the Spesh book :P

Thanks Hooligan :) I'm so lazy, I'll have to start it soon but I've got ages, especially if I take Gap year.

Hehehe.... Spesh always seems to be daunting, but actually is just a different part of Maths, and just so happens to be 'hard' as most perceive it. :P

After having a look at the Heinemann textbook, I think it's very physic-sy. :)

Ah yes, half of Unit 4 is very very physic-sy. xD